Problem: Solve for $x$ and $y$ using substitution. ${3x-2y = -12}$ ${y = 4x+1}$
Since $y$ has already been solved for, substitute $4x+1$ for $y$ in the first equation. ${3x - 2}{(4x+1)}{= -12}$ Simplify and solve for $x$ $3x-8x - 2 = -12$ $-5x-2 = -12$ $-5x-2{+2} = -12{+2}$ $-5x = -10$ $\dfrac{-5x}{{-5}} = \dfrac{-10}{{-5}}$ ${x = 2}$ Now that you know ${x = 2}$ , plug it back into $\thinspace {y = 4x+1}\thinspace$ to find $y$ ${y = 4}{(2)}{ + 1}$ $y = 8 + 1$ $y = 9$ You can also plug ${x = 2}$ into $\thinspace {3x-2y = -12}\thinspace$ and get the same answer for $y$ : ${3}{(2)}{ - 2y = -12}$ ${y = 9}$